Math Problem Statement

A 6-sided die with faces labeled 1 to 6 will be rolled once. The possible outcomes are listed below, and each outcome has the same probability. Calculate the probability for each of the following events: (a) Event A: Rolling an odd number. (b) Event B: Rolling a number less than 3. (c) Event A and B: Rolling an odd number and rolling a number less than 3. (d) Event A or B: Rolling an odd number or rolling a number less than 3. Complete each calculation and verify the equation P(A) + P(B) - P(A and B).

Solution

Let's analyze the problem based on the instructions provided in the image.

  1. Define Events and Probabilities:

    • Event A: Rolling an odd number.
    • Event B: Rolling a number less than 3.

    For each event, we need to:

    • Identify the outcomes that satisfy each event.
    • Calculate the probability of each event.

Step-by-Step Solutions

  1. Determine Outcomes for Each Event:

    • Event A: Rolling an odd number. Possible outcomes are 1, 3, and 5.

      • Probability of Event A, P(A)P(A): There are 3 favorable outcomes out of 6 possible outcomes, so P(A)=36=12P(A) = \frac{3}{6} = \frac{1}{2}.
    • Event B: Rolling a number less than 3. Possible outcomes are 1 and 2.

      • Probability of Event B, P(B)P(B): There are 2 favorable outcomes out of 6, so P(B)=26=13P(B) = \frac{2}{6} = \frac{1}{3}.
    • Event A and B (Intersection): Rolling an odd number that is also less than 3. This only occurs for the outcome 1.

      • Probability of ABA \cap B: There is 1 favorable outcome out of 6, so P(AB)=16P(A \cap B) = \frac{1}{6}.
    • Event A or B (Union): Rolling an odd number or a number less than 3. Possible outcomes are 1, 2, 3, and 5.

      • Probability of ABA \cup B: There are 4 favorable outcomes out of 6, so P(AB)=46=23P(A \cup B) = \frac{4}{6} = \frac{2}{3}.
  2. Calculate P(A)+P(B)P(AB)P(A) + P(B) - P(A \cap B):

    Substitute the values we’ve calculated: P(A)+P(B)P(AB)=12+1316P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6}

    To add these fractions, we can convert them to a common denominator: 12=36,13=26,16=16\frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6}, \quad \frac{1}{6} = \frac{1}{6} 36+2616=46=23\frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}

    So, P(A)+P(B)P(AB)=23P(A) + P(B) - P(A \cap B) = \frac{2}{3}.

  3. Verify the Equation in Part (c):

    Since we found that P(A)+P(B)P(AB)=23P(A) + P(B) - P(A \cap B) = \frac{2}{3}, select the option that matches this value.


Let me know if you need further explanation or have any questions!

Additional Questions for Practice

  1. If the die had 8 faces, how would the probabilities change for each event?
  2. What would P(AB)P(A \cup B) be if the events were mutually exclusive?
  3. Can you calculate the probability of not rolling an odd number?
  4. What is the probability of rolling a number greater than 4?
  5. How would probabilities change if the die was weighted?

Tip

When calculating probabilities for combined events, it helps to list all possible outcomes first, as it simplifies finding intersections and unions.

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Math Problem Analysis

Mathematical Concepts

Probability
Set Theory
Intersection and Union of Events

Formulas

P(A or B) = P(A) + P(B) - P(A and B)
Probability of an event = (Number of favorable outcomes) / (Total number of outcomes)

Theorems

Addition Rule for Probability

Suitable Grade Level

Grades 6-8